Problem: $f(x) = \begin{cases} 9\sqrt{x } & \text{for} ~~~~x\gt{0} \\ -2x& \text{for} ~~~~ x \leq0\end{cases}$ Evaluate the definite integral. $\int^1_{-3}f(x)\,dx = $ Choose 1 answer: Choose 1 answer: (Choice A) A $-12$ (Choice B) B $-9$ (Choice C) C $15$ (Choice D) D $16$
Answer: Splitting up the definite integral Since we're working with a piecewise function, we need to split the definite integral up into two pieces: $\phantom{=} \int^1_{-3}f(x)\,dx$ $= \int^1_{0}f(x)\,dx + \int^{0}_{-3}f(x)\,dx~~~~~~$ [Why did we split at 0?] $= \int^1_{0}9\sqrt{x}\,dx + \int^0_{-3}-2x\,dx ~~~~~~$ Evaluating each piece Next, let's evaluate each of these definite integrals one at a time. The first definite integral: $\begin{aligned} \int^1_{0}9\sqrt{x}\,dx &=6x^\frac32\Bigg|^1_{{0}} \\\\ &= \left[6 ( 1)^\frac32 \right] - \left[6 ({0})^\frac32\right] \\\\ &= \left[6\right] -\left[0 \right] \\\\ &= {6}\end{aligned}$ The second definite integral: $\begin{aligned} \int^0_{-3}-2x\,dx &=-x^2\Bigg|^0_{{-3}} \\\\ &= \left[- ( 0)^2 \right] - \left[- ({-3})^2\right] \\\\ &= \left[0\right] -\left[-9 \right] \\\\ &= {9}\end{aligned}$ Putting the pieces together Now let's add these two pieces together to find the answer: $\phantom{=} \int^1_{0}9\sqrt{x}\,dx + \int^0_{-3}-2x\,dx$ $ = {6} + {9}$ $ = 15$ The answer $\int^1_{-3}f(x)\,dx = 15$